0.000 000 000 000 268 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 268(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.000 000 000 000 268(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 268.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 268 × 2 = 0 + 0.000 000 000 000 536;
  • 2) 0.000 000 000 000 536 × 2 = 0 + 0.000 000 000 001 072;
  • 3) 0.000 000 000 001 072 × 2 = 0 + 0.000 000 000 002 144;
  • 4) 0.000 000 000 002 144 × 2 = 0 + 0.000 000 000 004 288;
  • 5) 0.000 000 000 004 288 × 2 = 0 + 0.000 000 000 008 576;
  • 6) 0.000 000 000 008 576 × 2 = 0 + 0.000 000 000 017 152;
  • 7) 0.000 000 000 017 152 × 2 = 0 + 0.000 000 000 034 304;
  • 8) 0.000 000 000 034 304 × 2 = 0 + 0.000 000 000 068 608;
  • 9) 0.000 000 000 068 608 × 2 = 0 + 0.000 000 000 137 216;
  • 10) 0.000 000 000 137 216 × 2 = 0 + 0.000 000 000 274 432;
  • 11) 0.000 000 000 274 432 × 2 = 0 + 0.000 000 000 548 864;
  • 12) 0.000 000 000 548 864 × 2 = 0 + 0.000 000 001 097 728;
  • 13) 0.000 000 001 097 728 × 2 = 0 + 0.000 000 002 195 456;
  • 14) 0.000 000 002 195 456 × 2 = 0 + 0.000 000 004 390 912;
  • 15) 0.000 000 004 390 912 × 2 = 0 + 0.000 000 008 781 824;
  • 16) 0.000 000 008 781 824 × 2 = 0 + 0.000 000 017 563 648;
  • 17) 0.000 000 017 563 648 × 2 = 0 + 0.000 000 035 127 296;
  • 18) 0.000 000 035 127 296 × 2 = 0 + 0.000 000 070 254 592;
  • 19) 0.000 000 070 254 592 × 2 = 0 + 0.000 000 140 509 184;
  • 20) 0.000 000 140 509 184 × 2 = 0 + 0.000 000 281 018 368;
  • 21) 0.000 000 281 018 368 × 2 = 0 + 0.000 000 562 036 736;
  • 22) 0.000 000 562 036 736 × 2 = 0 + 0.000 001 124 073 472;
  • 23) 0.000 001 124 073 472 × 2 = 0 + 0.000 002 248 146 944;
  • 24) 0.000 002 248 146 944 × 2 = 0 + 0.000 004 496 293 888;
  • 25) 0.000 004 496 293 888 × 2 = 0 + 0.000 008 992 587 776;
  • 26) 0.000 008 992 587 776 × 2 = 0 + 0.000 017 985 175 552;
  • 27) 0.000 017 985 175 552 × 2 = 0 + 0.000 035 970 351 104;
  • 28) 0.000 035 970 351 104 × 2 = 0 + 0.000 071 940 702 208;
  • 29) 0.000 071 940 702 208 × 2 = 0 + 0.000 143 881 404 416;
  • 30) 0.000 143 881 404 416 × 2 = 0 + 0.000 287 762 808 832;
  • 31) 0.000 287 762 808 832 × 2 = 0 + 0.000 575 525 617 664;
  • 32) 0.000 575 525 617 664 × 2 = 0 + 0.001 151 051 235 328;
  • 33) 0.001 151 051 235 328 × 2 = 0 + 0.002 302 102 470 656;
  • 34) 0.002 302 102 470 656 × 2 = 0 + 0.004 604 204 941 312;
  • 35) 0.004 604 204 941 312 × 2 = 0 + 0.009 208 409 882 624;
  • 36) 0.009 208 409 882 624 × 2 = 0 + 0.018 416 819 765 248;
  • 37) 0.018 416 819 765 248 × 2 = 0 + 0.036 833 639 530 496;
  • 38) 0.036 833 639 530 496 × 2 = 0 + 0.073 667 279 060 992;
  • 39) 0.073 667 279 060 992 × 2 = 0 + 0.147 334 558 121 984;
  • 40) 0.147 334 558 121 984 × 2 = 0 + 0.294 669 116 243 968;
  • 41) 0.294 669 116 243 968 × 2 = 0 + 0.589 338 232 487 936;
  • 42) 0.589 338 232 487 936 × 2 = 1 + 0.178 676 464 975 872;
  • 43) 0.178 676 464 975 872 × 2 = 0 + 0.357 352 929 951 744;
  • 44) 0.357 352 929 951 744 × 2 = 0 + 0.714 705 859 903 488;
  • 45) 0.714 705 859 903 488 × 2 = 1 + 0.429 411 719 806 976;
  • 46) 0.429 411 719 806 976 × 2 = 0 + 0.858 823 439 613 952;
  • 47) 0.858 823 439 613 952 × 2 = 1 + 0.717 646 879 227 904;
  • 48) 0.717 646 879 227 904 × 2 = 1 + 0.435 293 758 455 808;
  • 49) 0.435 293 758 455 808 × 2 = 0 + 0.870 587 516 911 616;
  • 50) 0.870 587 516 911 616 × 2 = 1 + 0.741 175 033 823 232;
  • 51) 0.741 175 033 823 232 × 2 = 1 + 0.482 350 067 646 464;
  • 52) 0.482 350 067 646 464 × 2 = 0 + 0.964 700 135 292 928;
  • 53) 0.964 700 135 292 928 × 2 = 1 + 0.929 400 270 585 856;
  • 54) 0.929 400 270 585 856 × 2 = 1 + 0.858 800 541 171 712;
  • 55) 0.858 800 541 171 712 × 2 = 1 + 0.717 601 082 343 424;
  • 56) 0.717 601 082 343 424 × 2 = 1 + 0.435 202 164 686 848;
  • 57) 0.435 202 164 686 848 × 2 = 0 + 0.870 404 329 373 696;
  • 58) 0.870 404 329 373 696 × 2 = 1 + 0.740 808 658 747 392;
  • 59) 0.740 808 658 747 392 × 2 = 1 + 0.481 617 317 494 784;
  • 60) 0.481 617 317 494 784 × 2 = 0 + 0.963 234 634 989 568;
  • 61) 0.963 234 634 989 568 × 2 = 1 + 0.926 469 269 979 136;
  • 62) 0.926 469 269 979 136 × 2 = 1 + 0.852 938 539 958 272;
  • 63) 0.852 938 539 958 272 × 2 = 1 + 0.705 877 079 916 544;
  • 64) 0.705 877 079 916 544 × 2 = 1 + 0.411 754 159 833 088;
  • 65) 0.411 754 159 833 088 × 2 = 0 + 0.823 508 319 666 176;
  • 66) 0.823 508 319 666 176 × 2 = 1 + 0.647 016 639 332 352;
  • 67) 0.647 016 639 332 352 × 2 = 1 + 0.294 033 278 664 704;
  • 68) 0.294 033 278 664 704 × 2 = 0 + 0.588 066 557 329 408;
  • 69) 0.588 066 557 329 408 × 2 = 1 + 0.176 133 114 658 816;
  • 70) 0.176 133 114 658 816 × 2 = 0 + 0.352 266 229 317 632;
  • 71) 0.352 266 229 317 632 × 2 = 0 + 0.704 532 458 635 264;
  • 72) 0.704 532 458 635 264 × 2 = 1 + 0.409 064 917 270 528;
  • 73) 0.409 064 917 270 528 × 2 = 0 + 0.818 129 834 541 056;
  • 74) 0.818 129 834 541 056 × 2 = 1 + 0.636 259 669 082 112;
  • 75) 0.636 259 669 082 112 × 2 = 1 + 0.272 519 338 164 224;
  • 76) 0.272 519 338 164 224 × 2 = 0 + 0.545 038 676 328 448;
  • 77) 0.545 038 676 328 448 × 2 = 1 + 0.090 077 352 656 896;
  • 78) 0.090 077 352 656 896 × 2 = 0 + 0.180 154 705 313 792;
  • 79) 0.180 154 705 313 792 × 2 = 0 + 0.360 309 410 627 584;
  • 80) 0.360 309 410 627 584 × 2 = 0 + 0.720 618 821 255 168;
  • 81) 0.720 618 821 255 168 × 2 = 1 + 0.441 237 642 510 336;
  • 82) 0.441 237 642 510 336 × 2 = 0 + 0.882 475 285 020 672;
  • 83) 0.882 475 285 020 672 × 2 = 1 + 0.764 950 570 041 344;
  • 84) 0.764 950 570 041 344 × 2 = 1 + 0.529 901 140 082 688;
  • 85) 0.529 901 140 082 688 × 2 = 1 + 0.059 802 280 165 376;
  • 86) 0.059 802 280 165 376 × 2 = 0 + 0.119 604 560 330 752;
  • 87) 0.119 604 560 330 752 × 2 = 0 + 0.239 209 120 661 504;
  • 88) 0.239 209 120 661 504 × 2 = 0 + 0.478 418 241 323 008;
  • 89) 0.478 418 241 323 008 × 2 = 0 + 0.956 836 482 646 016;
  • 90) 0.956 836 482 646 016 × 2 = 1 + 0.913 672 965 292 032;
  • 91) 0.913 672 965 292 032 × 2 = 1 + 0.827 345 930 584 064;
  • 92) 0.827 345 930 584 064 × 2 = 1 + 0.654 691 861 168 128;
  • 93) 0.654 691 861 168 128 × 2 = 1 + 0.309 383 722 336 256;
  • 94) 0.309 383 722 336 256 × 2 = 0 + 0.618 767 444 672 512;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 268(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011 0110 1111 0110 1111 0110 1001 0110 1000 1011 1000 0111 10(2)

5. Positive number before normalization:

0.000 000 000 000 268(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011 0110 1111 0110 1111 0110 1001 0110 1000 1011 1000 0111 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 42 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 268(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011 0110 1111 0110 1111 0110 1001 0110 1000 1011 1000 0111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011 0110 1111 0110 1111 0110 1001 0110 1000 1011 1000 0111 10(2) × 20 =

1.0010 1101 1011 1101 1011 1101 1010 0101 1010 0010 1110 0001 1110(2) × 2-42


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -42

Mantissa (not normalized):
1.0010 1101 1011 1101 1011 1101 1010 0101 1010 0010 1110 0001 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-42 + 2(11-1) - 1 =

(-42 + 1 023)(10) =

981(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 981 ÷ 2 = 490 + 1;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

981(10) =

011 1101 0101(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 1101 1011 1101 1011 1101 1010 0101 1010 0010 1110 0001 1110 =

0010 1101 1011 1101 1011 1101 1010 0101 1010 0010 1110 0001 1110


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1101 0101

Mantissa (52 bits) =
0010 1101 1011 1101 1011 1101 1010 0101 1010 0010 1110 0001 1110


Decimal number 0.000 000 000 000 268 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0101 - 0010 1101 1011 1101 1011 1101 1010 0101 1010 0010 1110 0001 1110

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100